Clock Error

Figure 3.8: Clock error.

Consider two clocks, $A$ and $B$, that are located a distance $l_0$ apart in their mutual rest frame. Suppose that the clocks are synchronized using light pulses emitted from a source that lies half-way between them. Let us observe the clocks in a reference frame that moves with velocity ${\bf v}$ with respect to the clocks' rest frame in a direction that is parallel to their mutual displacement. See Figure 3.8. In the moving frame, the contracted distance between the two clocks is $l_0/\gamma$, but the source is still located half-way between the clocks. Moreover, the two clocks appear to move with the same velocity, ${\bf -v}$. Consider a light pulse that is emitted by the source and travels to the two clocks. Suppose that, in the moving frame, it takes a time $t_a$ for the pulse to travel from the source to clock $A$. The pulse travels a distance $l_0/(2\,\gamma)+ v\,t_a$. Thus, given that the pulse travels at the speed $c$, according to Einstein's second postulate, we have

$$t_a = \frac{l_0/(2\,\gamma)+ v\,t_a}{c},$$ (3.82)

or

$$t_a = \frac{l_0}{2\,\gamma\,(c-v)}.$$ (3.83)

Suppose that, in the moving frame, it takes a time $t_b$ for the pulse to travel from the source to clock $B$. The pulse travels a distance $l_0/(2\,\gamma)- v\,t_b$. Thus, given that the pulse travels at the speed $c$, we have

$$t_b = \frac{l_0/(2\,\gamma)-v\,t_b}{c},$$ (3.84)

or

$$t_b = \frac{l_0}{2\,\gamma\,(c+v)}.$$ (3.85)

Now, in the clocks' rest frame, the pulse arrives at clocks $A$ and $B$ simultaneously. However, in the moving frame, the pulse arrives at clock $B$ prior to its arrival at clock $A$ (because $t_b<t_a$). In other words, two events, a spatial distance $l_0$ apart, that take place simultaneously in a particular reference frame, do not appear to take place simultaneously in a reference frame that moves with velocity ${\bf v}$ in the direction of the mutual displacement of the two events. This phenomenon is known as clock error. The time difference between the two events in the moving frame is

$${\mit\Delta} t= t_a-t_b = \frac{l_0}{2\,\gamma}\left(\frac{1}{c-v}-\frac{1}{c+v}\right)= \frac{l_0}{2\,\gamma}\frac{2\,v}{c^2-v^2}.$$ (3.86)

which reduces to

$${\mit\Delta t} = \frac{\gamma\,v\,l_0}{c^2}.$$ (3.87)